Certification Problem

Input (TPDB TRS_Standard/AG01/#3.1)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  quot#(s(x),s(y))

  →

  quot#(minus(x,y),s(y))

  (6)

  1.1.1.1 Usable Rules Processor

  We restrict the rewrite rules to the following usable rules of the DP problem.

  minus(x,0)	→	x	(1)
  minus(s(x),s(y))	→	minus(x,y)	(2)

  1.1.1.1.1 Innermost Lhss Removal Processor

  We restrict the innermost strategy to the following left hand sides.

  minus(x0,0)

  minus(s(x0),s(x1))

  1.1.1.1.1.1 Reduction Pair Processor

  Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

  prec(s)

  =

  0

  weight(s)

  =

  1

  in combination with the following argument filter

  π(quot#)	=	1
  π(s)	=	[1]
  π(minus)	=	1

  the pair

  quot#(s(x),s(y))

  →

  quot#(minus(x,y),s(y))

  (6)

  could be deleted.

  1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  minus#(s(x),s(y))

  →

  minus#(x,y)

  (5)

  1.1.1.2 Usable Rules Processor

  We restrict the rewrite rules to the following usable rules of the DP problem.

  There are no rules.

  1.1.1.2.1 Innermost Lhss Removal Processor

  We restrict the innermost strategy to the following left hand sides.

  There are no lhss.

  1.1.1.2.1.1 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  minus#(s(x),s(y))	→	minus#(x,y)	(5)
  1	>	1
  2	>	2

  As there is no critical graph in the transitive closure, there are no infinite chains.